1 Basic Considerations Outline 1.1 Introduction 1.2 Dimensions, Units, and Physical Quantities 1.3 Continuum View of Gases and Liquids 1.4 Pressure and Temperature Scales 1.5 Fluid Properties 1.5.1 Density and Specific Weight 1.5.2 Viscosity 1.5.3 Compressibility 1.5.4 Surface Tension 1.5.5 Vapor Pressure 1.6 Conservation Laws 1.7 Thermodynamic Properties and Relationships 1.7.1 Properties of an Ideal Gas 1.7.2 First Law of Thermodynamics 1.7.3 Other Thermodynamic Quantities 1.8 Summary Chapter Objectives The objectives of this chapter are to: Introduce many of the quantities encountered in fluid mechanics including their dimensions and units. Identify the liquids to be considered in this text. Introduce the fluid properties of interest. Present the thermodynamic laws and associated quantities. 3
4 Chapter 1 / Basic Considerations KEY CONCEPT We will present the fundamentals of fluids so that engineers are able to understand the role that fluid plays in particular applications. 1.1 INTRODUCTION A proper understanding of fluid mechanics is extremely important in many areas of engineering. In biomechanics the flow of blood and cerebral fluid are of particular interest; in meteorology and ocean engineering an understanding of the motions of air movements and ocean currents requires a knowledge of the mechanics of fluids, chemical engineers must understand fluid mechanics to design the many different kinds of chemical-processing equipment; aeronautical engineers use their knowledge of fluids to maximize lift and minimize drag on aircraft and to design fan-jet engines; mechanical engineers design pumps, turbines, internal combustion engines, air compressors, air-conditioning equipment, pollution-control equipment, and power plants using a proper understanding of fluid mechanics; and civil engineers must also utilize the results obtained from a study of the mechanics of fluids to understand the transport of river sediment and erosion, the pollution of the air and water, and to design piping systems, sewage treatment plants, irrigation channels, flood control systems, dams, and domed athletic stadiums. It is not possible to present fluid mechanics in such a way that all of the foregoing subjects can be treated specifically; it is possible, however, to present the fundamentals of the mechanics of fluids so that engineers are able to understand the role that the fluid plays in a particular application. This role may involve the proper sizing of a pump (the horsepower and flow rate) or the calculation of a force acting on a structure. In this book we present the general equations, both integral and differential, that result from the conservation of mass principle, Newton s second law, and the first law of thermodynamics. From these a number of particular situations will be considered that are of special interest. After studying this book the engineer should be able to apply the basic principles of the mechanics of fluids to new and different situations. In this chapter topics are presented that are directly or indirectly relevant to all subsequent chapters. We include a macroscopic description of fluids, fluid properties, physical laws dominating fluid mechanics, and a summary of units and dimensions of important physical quantities. Before we can discuss quantities of interest, we must present the units and dimensions that will be used in our study of fluid mechanics. 1.2 DIMENSIONS, UNITS, AND PHYSICAL QUANTITIES Before we begin the more detailed studies of fluid mechanics, let us discuss the dimensions and units that will be used throughout the book. Physical quantities require quantitative descriptions when solving an engineering problem. Density is one such physical quantity. It is a measure of the mass contained in a unit volume. Density does not, however, represent a fundamental dimension. There are nine quantities that are considered to be fundamental dimensions: length, mass, time, temperature, amount of a substance, electric current, luminous intensity, plane angle, and solid angle. The dimensions of all other quantities can be expressed in terms of the fundamental dimensions. For example, the quantity
Sec. 1.2 / Dimensions, Units, and Physical Quantities 5 force can be related to the fundamental dimensions of mass, length, and time. To do this, we use Newton s second law, named after Sir Isaac Newton (1642 1727), expressed in simplified form in one direction as F ma (1.2.1) Using brackets to denote the dimension of, this is written dimensionally as [F ] [m][a] F M T L 2 (1.2.2) where F, M, L, and T are the dimensions of force, mass, length, and time, respectively. If force had been selected as a fundamental dimension rather than mass, a common alternative, mass would have dimensions of [m] [ F] [ a] M F T 2 L (1.2.3) where F is the dimension 1 of force. There are also systems of dimensions in which both mass and force are selected as fundamental dimensions. In such systems conversion factors, such as a gravitational constant, are required; we do not consider these types of systems in this book, so they will not be discussed. To give the dimensions of a quantity a numerical value, a set of units must be selected. In the United States, two primary systems of units are presently being used, the British Gravitational System, which we will refer to as English units, and the International System, which is referred to as SI (Système International) units. SI units are preferred and are used internationally; the United States is the only major country not requiring the use of SI units, but there is now a program of conversion in most industries to the predominant use of SI units. Following this trend, we have used primarily SI units. However, as English units are still in use, some examples and problems are presented in these units as well. The fundamental dimensions and their units are presented in Table 1.1; some derived units appropriate to fluid mechanics are given in Table 1.2. Other units that are acceptable are the hectare (ha), which is 10 000 m 2, used for large areas; the metric ton (t), which is 1000 kg, used for large masses; and the liter (L), which is 0.001 m 3. Also, density is occasionally expressed as grams per liter (g/l). In chemical calculations the mole is often a more convenient unit than the kilogram. In some cases it is also useful in fluid mechanics. For gases the KEY CONCEPT SI units are preferred and are used internationally. 1 Unfortunately, the quantity force F and the dimension of force [F ] use the same symbol.
6 Chapter 1 / Basic Considerations Table 1.1 Fundamental Dimensions and Their Units Quantity Dimensions SI units English units Length L meter m foot ft Mass m M kilogram kg slug slug Time t T second s second sec Electric current i ampere A ampere A Temperature T kelvin K Rankine R Amount of substance M kg-mole kmol lb-mole lbmol Luminous intensity candela cd candela cd Plane angle radian rad radian rad Solid angle steradian sr steradian sr kilogram-mole (kmol) is the quantity that fills the same volume as 32 kilograms of oxygen at the same temperature and pressure.the mass (in kilograms) of a gas filling that volume is equal to the molecular weight of the gas; for example, the mass of 1 kmol of nitrogen is 28 kilograms. When expressing a quantity with a numerical value and a unit, prefixes have been defined so that the numerical value may be between 0.1 and 1000. These Table 1.2 Derived Units Quantity Dimensions SI units English units Area A L 2 m 2 ft 2 Volume V L 3 m 3 ft 3 L (liter) Velocity V L/T m/s ft/sec Acceleration a L/T 2 m/s 2 ft/sec 2 Angular velocity v T 1 rad/s rad/sec Force F ML/T 2 kgm/s 2 slug-ft/sec 2 N (newton) lb (pound) Density r M/L 3 kg/m 3 slug/ft 3 Specific weight g M/L 2 T 2 N/m 3 lb/ft 3 Frequency f T 1 s 1 sec 1 Pressure p M/LT 2 N/m 2 lb/ft 2 Pa (pascal) (psf) Stress t M/LT 2 N/m 2 lb/ft 2 Pa (pascal) (psf) Surface tension s M/T 2 N/m lb/ft Work W ML 2 /T 2 Nm ft-lb Energy E ML 2 /T 2 J (joule) Nm ft-lb J (joule) Heat rate Q ML 2 /T 3 J/s Btu/sec Torque T ML 2 /T 2 Nm ft-lb Power P ML 2 /T 3 J/s ft-lb/sec Ẇ W (watt) Viscosity m M/LT Ns/m 2 lb-sec/ft 2 Mass flux ṁ M/T kg/s slug/sec Flow rate Q L 3 /T m 3 /s ft 3 /sec Specific heat c L 2 /T 2 J/kgK Btu/slug- R Conductivity K ML/T 3 W/mK lb/sec- R
Sec. 1.2 / Dimensions, Units, and Physical Quantities 7 Table 1.3 SI Prefixes Multiplication factor Prefix Symbol 10 12 tera T 10 9 giga G 10 6 mega M 10 3 kilo k 10 2 centi a c 10 3 milli m 10 6 micro 10 9 nano n 10 12 pico p a Permissible if used alone as cm, cm 2, or cm 3. prefixes are presented in Table 1.3. Using scientific notation, however, we use powers of 10 rather than prefixes (e.g., 2 10 6 N rather than 2 MN). If larger numbers are written the comma is not used; twenty thousand would be written as 20 000 with a space and no comma. 2 Newton s second law relates a net force acting on a rigid body to its mass and acceleration. This is expressed as F ma (1.2.4) KEY CONCEPT When using SI units, if larger numbers are written (5 digits or more), the comma is not used. The comma is replaced by a space (i.e., 20 000). Consequently, the force needed to accelerate a mass of 1 kilogram at 1 meter per second squared in the direction of the net force is 1 newton; using English units, the force needed to accelerate a mass of 1 slug at 1 foot per second squared in the direction of the net force is 1 pound. This allows us to relate the units by N kgm/s 2 lb slug-ft/sec 2 (1.2.5) which are included in Table 1.2. These relationships between units are often used in the conversion of units. In the SI system, weight is always expressed in newtons, never in kilograms. In the English system, mass is usually expressed in slugs, although pounds are used in some thermodynamic relations. To relate weight to mass, we use KEY CONCEPT The relationship N kgm/s 2 is often used in the conversion of units. W mg (1.2.6) where g is the local gravity. The standard value for gravity is 9.80665 m/s 2 (32.174 ft/sec 2 ) and it varies from a minimum of 9.77 m/s 2 at the top of Mt. Everest to a maximum of 9.83 m/s 2 in the deepest ocean trench. A nominal value of 9.81 m/s 2 (32.2 ft/sec 2 ) will be used unless otherwise stated. Finally, a note on significant figures. In engineering calculations we often do not have confidence in a calculation beyond three significant digits since the information 2 In many countries commas represent decimal points, so they are not used where confusion may occur.
8 Chapter 1 / Basic Considerations KEY CONCEPT We will assume that all information given is known to three significant digits. given in the problem statement is often not known to more than three significant digits; in fact, viscosity and other fluid properties may not be known to even three significant digits.the diameter of a pipe may be stated as 2 cm; this would, in general, not be as precise as 2.000 cm would imply. If information used in the solution of a problem is known to only two significant digits, it is incorrect to express a result to more than two significant digits. In the examples and problems we will assume that all information given is known to three significant digits, and the results will be expressed accordingly. If the numeral 1 begins a number, it is not counted in the number of significant digits, i.e., the number 1.210 has three significant digits. Example 1.1 A mass of 100 kg is acted on by a 400-N force acting vertically upward and a 600-N force acting upward at a 45 angle. Calculate the vertical component of the acceleration. The local acceleration of gravity is 9.81 m/s 2. Solution The first step in solving a problem involving forces is to draw a free-body diagram with all forces acting on it, as shown in Fig. E1.1. 45 y 600 N W 400 N Fig. E1.1 Next, apply Newton s second law (Eq. 1.2.4). It relates the net force acting on a mass to the acceleration and is expressed as F y ma y Using the appropriate components in the y-direction, with W mg, we have 400 600 sin 45 100 9.81 100a y a y 1.567 m/s 2 The negative sign indicates that the acceleration is in the negative y-direction, i.e., down. Note: We have used only three significant digits in the answer since the information given in the problem is assumed known to three significant digits. (The number 1.567 has three significant digits. A leading 1 is not counted as a significant digit.) 1.3 CONTINUUM VIEW OF GASES AND LIQUIDS Substances referred to as fluids may be liquids or gases. In our study of the fluid mechanics we restrict the liquids that are studied. Before we state the restriction,
Sec. 1.3 / Continuum View of Gases and Liquids 9 we must define a shearing stress. A force F that acts on an area A can be decomposed into a normal component F n and a tangential component F t, as shown in Fig. 1.1. The force divided by the area upon which it acts is called a stress. The force vector divided by the area is a stress vector 3, the normal component of force divided by the area is a normal stress, and the tangential force divided by the area is a shear stress. In this discussion we are interested in the shear stress t. Mathematically, it is defined as t lim Ft (1.3.1) A 0 A Our restricted family of fluids may now be identified; the fluids considered in this book are those liquids and gases that move under the action of a shear stress, no matter how small that shear stress may be. This means that even a very small shear stress results in motion in the fluid. Gases obviously fall within this category of fluids, as do water and tar. Some substances, such as plastics and catsup, may resist small shear stresses without moving; a study of these substances is included in the subject of rheology and is not included in this book. It is worthwhile to consider the microscopic behavior of fluids in more detail. Consider the molecules of a gas in a container. These molecules are not stationary but move about in space with very high velocities. They collide with each other and strike the walls of the container in which they are confined, giving rise to the pressure exerted by the gas. If the volume of the container is increased while the temperature is maintained constant, the number of molecules impacting on a given area is decreased, and as a result, the pressure decreases. If the temperature of a gas in a given volume increases (i.e., the velocities of the molecules increase), the pressure increases due to increased molecular activity. Molecular forces in liquids are relatively high, as can be inferred from the following example. The pressure necessary to compress 20 grams of water vapor at 20 C into 20 cm 3, assuming that no molecular forces exist, can be shown by the ideal gas law to be approximately 1340 times the atmospheric pressure. Of course, this pressure is not required, because 20 g of water occupies 20 cm 3. It follows that the cohesive forces in the liquid phase must be very large. Despite the high molecular attractive forces in a liquid, some of the molecules at the surface escape into the space above. If the liquid is contained, an equilibrium is established between outgoing and incoming molecules. The presence of molecules above the liquid surface leads to a so-called vapor pressure. Stress vector: The force vector divided by the area. Normal stress: The normal component of force divided by the area. Shear stress: The tangential force divided by the area. Liquid: A state of matter in which the molecules are relatively free to change their positions with respect to each other but restricted by cohesive forces so as to maintain a relatively fixed volume. 4 Gas: A state of matter in which the molecules are practically unrestricted by cohesive forces. A gas has neither definite shape nor volume. KEY CONCEPT Fluids considered in this text are those that move under the action of a shear stress, no matter how small that stress may be. n ΔF Components ΔF n ΔA ΔA ΔF t Fig. 1.1 Normal and tangential components of a force. 3 A quantity that is defined in the margin is in bold face whereas a quantity not defined in the margin is italic. 4 Handbook of Chemistry and Physics, 40th ed. CRC Press, Boca Raton, Fla.
10 Chapter 1 / Basic Considerations Continuum: Continuous distribution of a liquid or gas throughout a region of interest. This pressure increases with temperature. For water at 20 C this pressure is approximately 0.02 times the atmospheric pressure. In our study of fluid mechanics it is convenient to assume that both gases and liquids are continuously distributed throughout a region of interest, that is, the fluid is treated as a continuum. The primary property used to determine if the continuum assumption is appropriate is the density r, defined by r lim m v 0 V (1.3.2) Standard atmospheric conditions: A pressure of 101.3 kpa and temperature of 15 C. KEY CONCEPT To determine if the continuum model is acceptable, compare a length l with the mean free path. Mean free path: The average distance a molecule travels before it collides with another molecule. where m is the incremental mass contained in the incremental volume V. The density for air at standard atmospheric conditions, that is, at a pressure of 101.3 kpa (14.7 psi) and a temperature of 15 C (59 F), is 1.23 kg/m 3 (0.00238 slug/ft 3 ). For water, the nominal value of density is 1000 kg/m 3 (1.94 slug/ft 3 ). Physically, we cannot let V 0 since, as V gets extremely small, the mass contained in V would vary discontinuously depending on the number of molecules in V; this is shown graphically in Fig. 1.2. Actually, the zero in the definition of density should be replaced by some small volume, below which the continuum assumption fails. For most engineering applications, the small volume shown in Fig. 1.2 is extremely small. For example, there are 2.7 10 16 molecules contained in a cubic millimeter of air at standard conditions; hence, is much smaller than a cubic millimeter. An appropriate way to determine if the continuum model is acceptable is to compare a characteristic length l (e.g., the diameter of a rocket) of the device or object of interest with the mean free path l, the average distance a molecule travels before it collides with another molecule; if l >> l, the continuum model is acceptable. The mean free path is derived in molecular theory. It is l 0.225 r m d 2 (1.3.3) where m is the mass (kg) of a molecule, r the density (kg/m 3 ) and d the diameter (m) of a molecule. For air m 4.8 10 26 kg and d 3.7 10 10 m. At standard atmospheric conditions the mean free path is approximately 6.4 10 6 cm, ρ ε ΔV Fig. 1.2 Density at a point in a continuum.
Sec. 1.4 / Pressure and Temperature Scales 11 at an elevation of 100 km it is 10 cm, and at 160 km it is 5000 cm. Obviously, at higher elevations the continuum assumption is not acceptable and the theory of rarefied gas dynamics (or free molecular flow) must be utilized. Satellites are able to orbit the earth if the primary dimension of the satellite is of the same order of magnitude as the mean free path. With the continuum assumption, fluid properties can be assumed to apply uniformly at all points in a region at any particular instant in time. For example, the density r can be defined at all points in the fluid; it may vary from point to point and from instant to instant; that is, in Cartesian coordinates r is a continuous function of x, y, z, and t, written as r (x, y, z, t). 1.4 PRESSURE AND TEMPERATURE SCALES In fluid mechanics pressure results from a normal compressive force acting on an area. The pressure p is defined as (see Fig. 1.3) p lim Fn A 0 A (1.4.1) where F n is the incremental normal compressive force acting on the incremental area A. The metric units to be used on pressure are newtons per square meter (N/m 2 ) or pascal (Pa). Since the pascal is a very small unit of pressure, it is more conventional to express pressure in units of kilopascal (kpa). For example, standard atmospheric pressure at sea level is 101.3 kpa. The English units for pressure are pounds per square inch (psi) or pounds per square foot (psf). Atmospheric pressure is often expressed as inches of mercury or feet of water, as shown in Fig. 1.4; such a column of fluid creates the pressure at the bottom of the column, providing the column is open to atmospheric pressure at the top. Both pressure and temperature are physical quantities that can be measured using different scales. There exist absolute scales for pressure and temperature, and there are scales that measure these quantities relative to selected reference points. In many thermodynamic relationships (see Section 1.7) absolute scales must be used for pressure and temperature. Figures 1.4 and 1.5 summarize the commonly used scales. The absolute pressure reaches zero when an ideal vacuum is achieved, that is, when no molecules are left in a space; consequently, a negative absolute pressure is an impossibility. A second scale is defined by measuring pressures relative KEY CONCEPT In many relationships, absolute scales must be used for pressure and temperature. Absolute pressure: The scale measuring pressure, where zero is reached when an ideal vacuum is achieved. ΔF n ΔA Surface Fig. 1.3 Definition of pressure.
12 Chapter 1 / Basic Considerations A A Positive pressure p A gage B Negative pressure or positive vacuum Standard atmosphere Local atmosphere p = 0 gage 101.3 kpa 14.7 psi 2117 psf 30.0 in. Hg 760 mm Hg 34 ft H 2 O 1.013 bar p A absolute B p B gage (negative) p B absolute Zero absolute pressure p = 0 absolute Fig. 1.4 Gage pressure and absolute pressure. Gage pressure: The scale measuring pressure relative to the local atmospheric pressure. to the local atmospheric pressure. This pressure is called gage pressure. A conversion from gage pressure to absolute pressure can be carried out using p absolute p atmospheric p gage (1.4.2) KEY CONCEPT Whenever the absolute pressure is less than the atmospheric pressure, it may be called a vacuum. Vacuum: Whenever the absolute pressure is less than the atmospheric pressure. Note that the atmospheric pressure in Eq. 1.4.2 is the local atmospheric pressure, which may change with time, particularly when a weather front moves through. However, if the local atmospheric pressure is not given, we use the value given for a particular elevation, as given in Table B.3 of Appendix B, and assume zero elevation if the elevation is unknown. The gage pressure is negative whenever the absolute pressure is less than atmospheric pressure; it may then be called a vacuum. In this book the word absolute will generally follow the pressure value if the pressure is given as an absolute pressure (e.g., p 50 kpa absolute). If it were stated as p 50 kpa, the pressure would be taken as a gage pressure, except that atmospheric pressure is always an absolute pressure. Most often in fluid mechanics gage pressure is used. C K F R Steam point Ice point Special point 100 373 212 672 0 273 32 492 18 255 0 460 Zero absolute temperature Fig. 1.5 Temperature scales.
Sec. 1.4 / Pressure and Temperature Scales 13 Two temperature scales are commonly used, the Celsius (C) and Fahrenheit (F) scales. Both scales are based on the ice point and steam point of water at an atmospheric pressure of 101.3 kpa (14.7 psi). Figure 1.5 shows that the ice and steam point are 0 and 100 C on the Celsius scale and 32 and 212 F on the Fahrenheit scale. There are two corresponding absolute temperature scales. The absolute scale corresponding to the Celsius scale is the kelvin (K) scale. The relation between these scales is K C 273.15 (1.4.3) The absolute scale corresponding to the Fahrenheit scale is the Rankine scale ( R). The relation between these scales is R F 459.67 (1.4.4) Note that in the SI system we do not write 100 K but simply 100 K, which is read 100 kelvins, similar to other units. Reference will often be made to standard atmospheric conditions or standard temperature and pressure. This refers to sea-level conditions at 40 latitude, which are taken to be 101.3 kpa (14.7 psi) for pressure and 15 C (59 F) for temperature. Actually, The standard pressure is usually taken as 100 kpa, sufficiently accurate for engineering calculations. KEY CONCEPT In the SI system, we write 100 K, which is read 100 kelvins. Example 1.2 A pressure gage attached to a rigid tank measures a vacuum of 42 kpa inside the tank shown in Fig. E1.2, which is situated at a site in Colorado where the elevation is 2000 m. Determine the absolute pressure inside the tank. air Fig. E1.2 42 kpa Solution To determine the absolute pressure, the atmospheric pressure must be known. If the elevation were not given, we would assume a standard atmospheric pressure of 100 kpa. However, with the elevation given, the atmospheric pressure is found from Table B.3 in Appendix B to be 79.5 kpa. Thus p 42 79.5 37.5 kpa absolute Note: A vacuum is always a negative gage pressure. Also, using standard atmospheric pressure of 100 kpa is acceptable, rather than 101.3 kpa, since it is within 1%, which is acceptable engineering accuracy.
14 Chapter 1 / Basic Considerations 1.5 FLUID PROPERTIES In this section we present several of the more common fluid properties. If density variation or heat transfer is significant, several additional properties, not presented here, become important. Specific weight: Weight per unit volume (g rg). 1.5.1 Density and Specific Weight Fluid density was defined in Eq. 1.3.2 as mass per unit volume. A fluid property directly related to density is the specific weight g or weight per unit volume. It is defined by g W V mg V rg (1.5.1) Specific gravity: The ratio of the density of a substance to the density of water. where g is the local gravity. The units of specific weight are N/m 3 (lb/ft 3 ). For water we use the nominal value of 9800 N/m 3 (62.4 lb/ft 3 ). The specific gravity S is often used to determine the specific weight or density of a fluid (usually a liquid). It is defined as the ratio of the density of a substance to the density of water at a reference temperature of 4 C: r g S rw gw ater ater (1.5.2) KEY CONCEPT Specific gravity is often used to determine the density of a fluid. For example, the specific gravity of mercury is 13.6, a dimensionless number; that is, the mass of mercury is 13.6 times that of water for the same volume. The density, specific weight, and specific gravity of air and water at standard conditions are given in Table 1.4. The density and specific weight of water do vary slightly with temperature; the approximate relationships are r H2 O 1000 (T 4) 2 180 g H2 O 9800 (T 4) 2 18 (1.5.3) TABLE 1.4 Density, Specific Weight, and Specific Gravity of Air and Water at Standard Conditions Density r Specific weight g kg/m 3 slug/ft 3 N/m 3 lb/ft 3 Specific gravity S Air 1.23 0.0024 12.1 0.077 0.00123 Water 1000 1.94 9810 62.4 1
Sec. 1.5 / Fluid Properties 15 For mercury the specific gravity relates to temperature by S Hg 13.6 0.0024T (1.5.4) Temperature in the three equations above is measured in degrees Celsius. For temperatures under 50 C, using the nominal values stated earlier for water and mercury, the error is less than 1%, certainly within engineering limits for most design problems. Note that the density of water at 0 C (32 F) is less than that at 4 C; consequently, the lighter water at 0 C rises to the top of a lake so that ice forms on the surface. For most other liquids the density at freezing is greater than the density just above freezing. 1.5.2 Viscosity Viscosity can be thought of as the internal stickiness of a fluid. It is one of the properties that influences the power needed to move an airfoil through the atmosphere. It accounts for the energy losses associated with the transport of fluids in ducts, channels, and pipes. Further, viscosity plays a primary role in the generation of turbulence. Needless to say, viscosity is an extremely important fluid property in our study of fluid flows. The rate of deformation of a fluid is directly linked to the viscosity of the fluid. For a given stress, a highly viscous fluid deforms at a slower rate than a fluid with a low viscosity. Consider the flow of Fig. 1.6 in which the fluid particles move in the x-direction at different speeds, so that particle velocities u vary with the y- coordinate. Two particle positions are shown at different times; observe how the particles move relative to one another. For such a simple flow field, in which u u(y), we can define the viscosity m of the fluid by the relationship Viscosity: The internal stickiness of a fluid. KEY CONCEPT Viscosity plays a primary role in the generation of turbulence. t m d u dy (1.5.5) where t is the shear stress of Eq. 1.3.1 and u is the velocity in the x-direction. The units of t are N/m 2 or Pa (lb/ft 2 ), and of m are Ns/m 2 (lb-sec/ft 2 ).The quantity du/dy is a velocity gradient and can be interpreted as a strain rate. Stress velocity-gradient relationships for more complicated flow situations are presented in Chapter 5. The concept of viscosity and velocity gradients can also be illustrated by considering a fluid within the small gap between two concentric cylinders, as shown Strain rate: The rate at which a fluid element deforms. y u(y) Particle 1 Particle 2 X X X t = 0 t = t 1 t = 2t 1 t = 3t 1 Fig. 1.6 Relative movement of two fluid particles in the presence of shear stresses.
16 Chapter 1 / Basic Considerations L u h ω r R u (a) (b) τ Rω u T (d) R r = R (c) r = R + h r Fig. 1.7 Fluid being sheared between cylinders with a small gap: (a) the two cylinders; (b) rotating inner cylinder; (c) velocity distribution; (d) the inner cylinder. The outer cylinder is fixed and the inner cylinder is rotating. in Fig. 1.7. A torque is necessary to rotate the inner cylinder at constant rotational speed v while the outer cylinder remains stationary. This resistance to the rotation of the cylinder is due to viscosity. The only stress that exists to resist the applied torque for this simple flow is a shear stress, which is observed to depend directly on the velocity gradient; that is, t m d u d (1.5.6) r where du/dr is the velocity gradient and u is the tangential velocity component, which depends only on r. For a small gap (h << R), this gradient can be approximated by assuming a linear velocity distribution 5 in the gap. Thus d u d r v (1.5.7) hr where h is the gap width. We can thus relate the applied torque T to the viscosity and other parameters by the equation T stress area moment arm t 2pRL R m v R 2pR 3 vlm 2pRL R h h (1.5.8) 5 If the gap is not small relative to R, the velocity distribution will not be linear (see Section 7.5). The distribution will also not be linear for relatively small values of v.
Sec. 1.5 / Fluid Properties 17 Stress τ Ideal plastic Non-Newtonian fluid (dilatant) Newtonian fluid Non-Newtonian fluid (pseudoplastic) Strain rate du/dy Fig. 1.8 Newtonian and non-newtonian fluids. where the shearing stress acting on the ends of the cylinder is negligible; L represents the length of the rotating cylinder. Note that the torque depends directly on the viscosity; thus the cylinders could be used as a viscometer, a device that measures the viscosity of a fluid. If the shear stress of a fluid is directly proportional to the velocity gradient, as was assumed in Eqs. 1.5.5 and 1.5.6, the fluid is said to be a Newtonian fluid. Fortunately, many common fluids, such as air, water, and oil, are Newtonian. Non-Newtonian fluids, with shear stress versus strain rate relationships as shown in Fig. 1.8, often have a complex molecular composition. Dilatants (quicksand, slurries) become more resistant to motion as the strain rate increases, and pseudoplastics (paint and catsup) become less resistant to motion with increased strain rate. Ideal plastics (or Bingham fluids) require a minimum shear stress to cause motion. Clay suspensions and toothpaste are examples that also require a minimum shear to cause motion, but they do not have a linear stress-strain rate relationship. An extremely important effect of viscosity is to cause the fluid to adhere to the surface; this is known as the no-slip condition. This was assumed in the example of Fig. 1.7. The velocity of the fluid at the rotating cylinder was taken to be vr, and the velocity of the fluid at the stationary cylinder was set equal to zero, as shown in Fig. 1.7b. When a space vehicle reenters the atmosphere, the high speed creates very large velocity gradients at the surface of the vehicle, resulting in large stresses that heat up the surface; the high temperatures can cause the vehicle to disintegrate if not properly protected. The viscosity is very dependent on temperature in liquids in which cohesive forces play a dominant role; note that the viscosity of a liquid decreases with increased temperature, as shown in Fig. B.1 in Appendix B. The curves are often approximated by the equation m Ae Bt (1.5.9) known as Andrade s equation; the constants A and B would be determined from measured data. For a gas it is molecular collisions that provide the internal stresses, so that as the temperature increases, resulting in increased molecular Viscosity 6, 454 Newtonian fluid: the shear stress of the fluid is directly proportional to the velocity gradient. KEY CONCEPT Viscosity causes fluid to adhere to a surface. No-slip condition: Condition where viscosity causes fluid to adhere to the surface. 6 To view a file on a specified page, simply open any of the eight major headings, then enter a page number in the box at the top and click on go to page. The numbers after the descriptors refer to the pages on the DVD.
18 Chapter 1 / Basic Considerations activity, the viscosity increases. This can be observed in the bottom curve for a gas of Fig. B.1 in Appendix B. Note, however, that the percentage change of viscosity in a liquid is much greater than in a gas for the same temperature difference. Also, one can show that cohesive forces and molecular activity are quite insensitive to pressure, so that m m(t) only for both liquids and gases. Since the viscosity is often divided by the density in the derivation of equations, it has become useful and customary to define kinematic viscosity to be v m r (1.5.10) where the units of v are m 2 /s (ft 2 /sec). Note that for a gas, the kinematic viscosity will also depend on the pressure since density is pressure sensitive. The kinematic viscosity is shown, at atmospheric pressure, in Fig. B.2 in Appendix B. Example 1.3 A viscometer is constructed with two 30-cm-long concentric cylinders, one 20.0 cm in diameter and the other 20.2 cm in diameter. A torque of 0.13 Nm is required to rotate the inner cylinder at 400 rpm (revolutions per minute). Calculate the viscosity. Solution The applied torque is just balanced by a resisting torque due to the shear stresses (see Fig. 1.7c). This is expressed by the small gap equation, Eq. 1.5.8. The radius is R d/2 10 cm; the gap h (d 2 d 1 ) 2 0.1 cm; the rotational speed, expressed as rad/s, is v 400 2p 60 41.89 rad/s. Equation 1.5.8 provides: Th m 2pR 3 vl 0.13(0.001) 0.001646 Ns m 2 2p(0.1) 3 (41.89)(0.3) Note: All lengths are in meters so that the desired units on m are obtained. The units can be checked by substitution: Nmm [m] Ns m 3 m 2 (rad/s)m Bulk modulus of elasticity: The ratio of change in pressure to relative change in density. 1.5.3 Compressibility In the preceding section we discussed the deformation of fluids that results from shear stresses. In this section we discuss the deformation that results from pressure changes. All fluids compress if the pressure increases, resulting in a decrease in volume or an increase in density. A common way to describe the compressibility of a fluid is by the following definition of the bulk modulus of elasticity B:
Sec. 1.5 / Fluid Properties 19 B lim V 0 p V lim V p T r 0 r r p V V r T p r T T (1.5.11) In words, the bulk modulus, also called the coefficient of compressibility, is defined as the ratio of the change in pressure (p) to relative change in density (r/r) while the temperature remains constant. The bulk modulus has the same units as pressure. The bulk modulus for water at standard conditions is approximately 2100 MPa (310,000 psi), or 21 000 times the atmospheric pressure. For air at standard conditions, B is equal to 1 atm. In general, B for a gas is equal to the pressure of the gas. To cause a 1% change in the density of water a pressure of 21 MPa (210 atm) is required. This is an extremely large pressure needed to cause such a small change; thus liquids are often assumed to be incompressible. For gases, if significant changes in density occur, say 4%, they should be considered as compressible; for small density changes under 3% they may also be treated as incompressible. This occurs for atmospheric airspeeds under about 100 m/s (220 mph), which includes many airflows of engineering interest: air flow around automobiles, landing and take-off of aircraft, and air flow in and around buildings. Small density changes in liquids can be very significant when large pressure changes are present. For example, they account for water hammer, which can be heard shortly after the sudden closing of a valve in a pipeline; when the valve is closed an internal pressure wave propagates down the pipe, producing a hammering sound due to pipe motion when the wave reflects from the closed valve or pipe elbows. Water hammer is considered in detail in Section 11.5. The bulk modulus can also be used to calculate the speed of sound in a liquid; in Section 9.2 it will be shown to be given by KEY CONCEPT Gases with small density changes under 3% may be treated as incompressible. c p r T B r (1.5.12) This yields approximately 1450 m/s (4800 ft/sec) for the speed of sound in water at standard conditions. The speed of sound in a gas will be presented in Section 1.7.3. 1.5.4 Surface Tension Surface tension is a property that results from the attractive forces between molecules. As such, it manifests itself only in liquids at an interface, usually a liquid-gas interface. The forces between molecules in the bulk of a liquid are equal in all directions, and as a result, no net force is exerted on the molecules. However, at an interface the molecules exert a force that has a resultant in the interface layer. This force holds a drop of water suspended on a rod and limits the size of the drop that may be held. It also causes the small drops from a Surface tension: A property resulting from the attractive forces between molecules.
20 Chapter 1 / Basic Considerations 2 πrσ 2 2 πrσ p πr 2 p πr 2 (a) Fig. 1.9 Internal forces in (a) a droplet and (b) a bubble. (b) KEY CONCEPT Force due to surface tension results from a length multiplied by the surface tension. Drop Formation, 453 sprayer or atomizer to assume spherical shapes. It may also play a significant role when two immiscible liquids (e.g., oil and water) are in contact with each other. Surface tension has units of force per unit length, N/m (lb/ft). The force due to surface tension results from a length multiplied by the surface tension; the length to use is the length of fluid in contact with a solid, or the circumference in the case of a bubble.a surface tension effect can be illustrated by considering the free-body diagrams of half a droplet and half a bubble, as shown in Fig. 1.9. The droplet has one surface, and the bubble is composed of a thin film of liquid with an inside surface and an outside surface. An expression for the pressure inside the droplet and bubble can now be derived. The pressure force ppr 2 in the droplet balances the surface tension force around the circumference. Hence ppr 2 2pRs p 2 s R (1.5.13) Similarly, the pressure force in the bubble is balanced by the surface tension forces on the two circumferences assuming the bubble thickness is small. Therefore, ppr 2 2(2pRs) p 4 s R (1.5.14) From Eqs. 1.5.13 and 1.5.14 we can conclude that the internal pressure in a bubble is twice as large as that in a droplet of the same size. Figure 1.10 shows the rise of a liquid in a clean glass capillary tube due to surface tension. The liquid makes a contact angle b with the glass tube. Experiments have shown that this angle for water and most liquids in a clean glass tube is zero. There are also cases for which this angle is greater than 90 (e.g., mercury); such liquids have a capillary drop. If h is the capillary rise, D the diameter, r the density, and s the surface tension, h can be determined from
Sec. 1.5 / Fluid Properties 21 β π Dσ Weight of water W h D Air Liquid Fig. 1.10 Rise in a capillary tube. equating the vertical component of the surface tension force to the weight of the liquid column: or, rearranged, 2 spd cos b g pd h (1.5.15) 4 h 4s cos b gd (1.5.16) Surface tension may influence engineering problems when, for example, laboratory modeling of waves is conducted at a scale that surface tension forces are of the same order of magnitude as gravitational forces. Example 1.4 A 2-mm-diameter clean glass tube is inserted in water at 15 C (Fig. E1.4). Determine the height that the water will climb up the tube. The water makes a contact angle of 0 with the clean glass. Capillary, 346 π Dσ h W = γ V D Air Water Fig. E1.4
22 Chapter 1 / Basic Considerations Solution A free-body diagram of the water shows that the upward surface-tension force is equal and opposite to the weight. Writing the surface-tension force as surface tension times distance, we have or spd g pd h 4 4s 4 0.0741 N/m h 0.01512 m or 15.12 mm g D 9800 N/m 3 0.002 m The numerical values for s and r were obtained from Table B.1 in Appendix B. Note that the nominal value used for the specific weight of water is g rg 9800 N/m 3. 2 Example 1.4a on the DVD, Similarity and Scaling, Capillary Rise 512 Fig. 1.11 Cooking food in boiling water takes a longer amount of time at a high altitude. It would take longer to boil these eggs in Denver than in New York City. (Thomas Firak Photography/FoodPix/ Getty Images) Vapor pressure: The pressure resulting from molecules in a gaseous state. KEY CONCEPT Cavitation can be very damaging. Boiling: The point where vapor pressure is equal to the atmospheric pressure. Cavitation: Bubbles form in a liquid when the local pressure falls below the vapor pressure of the liquid. 1.5.5 Vapor Pressure When a small quantity of liquid is placed in a closed container, a certain fraction of the liquid will vaporize. Vaporization will terminate when equilibrium is reached between the liquid and gaseous states of the substance in the container in other words, when the number of molecules escaping from the water surface is equal to the number of incoming molecules. The pressure resulting from molecules in the gaseous state is the vapor pressure. The vapor pressure is different from one liquid to another. For example, the vapor pressure of water at 15 C is 1.70 kpa absolute,and for ammonia it is 33.8 kpa absolute. The vapor pressure is highly dependent on temperature; it increases significantly when the temperature increases. For example, the vapor pressure of water increases to 101.3 kpa (14.7 psi) if the temperature reaches 100 C (212 F). Water vapor pressures for other temperatures are given in Appendix B. It is, of course, no coincidence that the water vapor pressure at 100 C is equal to the standard atmospheric pressure. At that temperature the water is boiling; that is, the liquid state of the water can no longer be sustained because the attractive forces are not sufficient to contain the molecules in a liquid phase. In general, a transition from the liquid state to the gaseous state occurs if the local absolute pressure is less than the vapor pressure of the liquid. At high elevations where the atmospheric pressure is relatively low, boiling occurs at temperatures less than 100 C; see Fig. 1.11. At an elevation of 3000 m, boiling would occur at approximately 90 C; see Tables B.3 and B.1. In liquid flows, conditions can be created that lead to a pressure below the vapor pressure of the liquid. When this happens, bubbles are formed locally. This phenomenon, called cavitation, can be very damaging when these bubbles are transported by the flow to higher-pressure regions. What happens is that the bubbles collapse upon entering the higher-pressure region, and this collapse produces local pressure spikes which have the potential of damaging a pipe wall or a ship s propeller. Cavitation on a propeller is shown in Fig. 1.12. Additional information on cavitation is included in Section 8.3.4.
Sec. 1.6 / Conservation Laws 23 Fig. 1.12 A photograph of a cavitating propeller inside MIT s water tunnel. (Courtesy of Prof. S. A. Kinnas, Ocean Engineering Group, University of Texas - Austin.) Example 1.5 Calculate the vacuum necessary to cause cavitation in a water flow at a temperature of 80 C in Colorado where the elevation is 2500 m. Solution The vapor pressure of water at 80 C is given in Table B.1. It is 47.3 kpa absolute. The atmospheric pressure is found by interpolation using Table B.3 to be 79.48 (79.48 61.64)500\2000 75.0. The required pressure is then p 47.3 75.0 27.7 kpa or 27.7 kpa vacuum 1.6 CONSERVATION LAWS From experience it has been found that fundamental laws exist that appear exact; that is, if experiments are conducted with the utmost precision and care, deviations from these laws are very small and in fact, the deviations would be even smaller if improved experimental techniques were employed. Three such laws form the basis for our study of fluid mechanics. The first is the conservation of mass, which states that matter is indestructible. Even though Einstein s theory of relativity postulates that under certain conditions, matter is convertible into energy and leads to the statement that the extraordinary quantities of radiation from the sun are associated with a conversion of 3.3 10 14 kg of matter per day into energy, the destructibility of matter under typical engineering conditions is not measurable and does not violate the conservation of mass principle. For the second and third laws it is necessary to introduce the concept of a system. A system is defined as a fixed quantity of matter upon which attention is focused. Everything external to the system is separated by the system boundaries. Conservation of mass: Matter is indestructible. System: matter. A fixed quantity of
24 Chapter 1 / Basic Considerations Newton s second law: The sum of all external forces acting on a system is equal to the time rate of change of linear momentum of the system. Conservation of energy: The total energy of an isolated system remains constant. Also known as the first law of thermodynamics. These boundaries may be fixed or movable, real, or imagined. With this definition we can now present our second fundamental law, the conservation of momentum: The momentum of a system remains constant if no external forces are acting on the system. A more specific law based on this principle is Newton s second law: The sum of all external forces acting on a system is equal to the time rate of change of linear momentum of the system. A parallel law exists for the moment of momentum:the rate of change of angular momentum is equal to the sum of all torques acting on the system. The third fundamental law is the conservation of energy, which is also known as the first law of thermodynamics: The total energy of an isolated system remains constant. If a system is in contact with the surroundings, its energy increases only if the energy of the surroundings experiences a corresponding decrease. It is noted that the total energy consists of potential, kinetic, and internal energy, the latter being the energy content due to the temperature of the system. Other forms of energy 7 are not considered in fluid mechanics. The first law of thermodynamics and other thermodynamic relationships are presented in the following section. Extensive property: A property that depends on the system s mass. Intensive property: A property that is independent of the system s mass. 1.7 THERMODYNAMIC PROPERTIES AND RELATIONSHIPS For incompressible fluids, the three laws mentioned in the preceding section suffice. This is usually true for liquids but also for gases if relatively small pressure, density, and temperature changes occur. However, for a compressible fluid, it may be necessary to introduce other relationships, so that density, temperature, and pressure changes are properly taken into account. An example is the prediction of changes in density, pressure, and temperature when compressed gas is released from a rocket through a nozzle. Thermodynamic properties, quantities that define the state of a system, either depend on the system s mass or are independent of the mass. The former is called an extensive property, and the latter is called an intensive property. An intensive property can be obtained by dividing the extensive property by the mass of the system. Temperature and pressure are intensive properties; momentum and energy are extensive properties. 1.7.1 Properties of an Ideal Gas The behavior of gases in most engineering applications can be described by the ideal-gas law, also called the perfect-gas law. When the temperature is relatively low and/or the pressure relatively high, caution should be exercised and real-gas laws should be applied. For air with temperatures higher than 50 C (58 F) the ideal-gas law approximates the behavior of air to an acceptable degree provided that the pressure is not extremely high. 7 Other forms of energy include electric and magnetic field energy, the energy associated with atoms, and energy released during combustion.
Sec. 1.7 / Thermodynamic Properties and Relationships 25 The ideal-gas law is given by p rrt (1.7.1) where p is the absolute pressure, r the density, T the absolute temperature, and R the gas constant. The gas constant is related to the universal gas constant R u by the relationship R R u (1.7.2) M where M is the molar mass. Values of M and R are tabulated in Table B.4 in Appendix B. The value of R u is R u 8.314 kj/kmolk 49,710 ft-lb/slugmol- R (1.7.3) For air M 28.97 kg/kmol (28.97 slug/slugmol), so that for air R 0.287 kj/kgk (1716 ft-lb/slug- R), a value used extensively in calculations involving air. Other forms that the ideal-gas law takes are pv mrt (1.7.4) and pv nr u T (1.7.5) where n is the number of moles. Example 1.6 A tank with a volume of 0.2 m 3 contains 0.5 kg of nitrogen. The temperature is 20 C. What is the pressure? Solution Assume this is an ideal gas. Apply Eq. 1.7.1 (R can be found in Table B.4). Solving the equation, p rrt, we obtain, using r = m/v, 0. 5 kg kj p 0 3 0.2968 (20 273) K 218 kpa absolute. 2 m kg K Note: The resulting units are kj/m 3 knm/m 3 kn/m 2 kpa. The ideal-gas law requires that pressure and temperature be in absolute units.